Find Pk₊₁ for the given P_k. $P _ { k } = \frac { 3 } { k ( k + 7 ) }$ A) $P _ { k + 1 } = \frac { 3 } { k ( k + 7 ) } + 1$ B) $P _ { k + 1 } = \frac { 3 } { k ( k + 7 ) } + \frac { 3 } { ( k + 1 ) ( k + 8 ) }$ C) $P _ { k + 1 } = \frac { 3 } { ( k + 1 ) ( k + 8 ) }$ D) $P _ { k + 1 } = \frac { 3 } { k ( k + 8 ) }$ E) $P _ { k + 1 } = \frac { 8 } { ( k + 1 ) ( k + 7 ) }$

Question

Quizplus · Accepted Answer

The expression for $P_{k+1}$ is obtained by substituting $k+1$ into the formula for $P_k$, resulting in $\frac{3}{(k+1)((k+1)+7)} = \frac{3}{(k+1)(k+8)}$.

Find Pk+1 for the Given Pk Pk=3k(k+7)P _ { k } = \frac { 3 } { k ( k + 7 ) }Pk​=k(k+7)3​

Find Pk₊₁ for the Given P_k $P _ { k } = \frac { 3 } { k ( k + 7 ) }$